Question

) an airline sold 560 tickets for an airbus 380 flight (capacity: 555 seats) in the assumption that not all passengers that bought a ticket will arrive for the flight. assume that the probability that a passenger will not shop up for the flight is 1%, independently for all passengers. how likely is it that there are more passengers showing up for the flight than seats are available? calculate this probability by using a normal approximation for the number of passengers that showed up for the flight, with and without applying the continuity correction.

Answer 1

This is a statistical probability problem that requires the use of the normal distribution. We can use the normal approximation to estimate the number of passengers that actually show up for the flight.

Let X be the number of passengers that show up for the flight. Since there are 560 tickets sold and the probability of a passenger not showing up is 1%, we have that the expected value of X is E(X) = 560 * (1 - 0.01) = 554.4 and the variance of X is Var(X) = 560 * 0.01 * 0.99 = 5.544.

To calculate the probability that there are more passengers showing up for the flight than seats available, we need to calculate the probability that X is greater than 555.

Using the normal distribution, we can calculate this probability as follows:

Z = (555 - 554.4) / sqrt(5.544) = 0.436

Using a normal distribution table or calculator, we can find the probability that Z is greater than 0.436 as P(Z > 0.436) = 1 - P(Z < 0.436) = 1 - 0.6664 = 0.3336.

Therefore, the probability that there are more passengers showing up for the flight than seats available is approximately 0.3336.

To apply the continuity correction, we can adjust the value of 555 to 555.5.

Z = (555.5 - 554.4) / sqrt(5.544) = 0.710

Using a normal distribution table or calculator, we can find the probability that Z is greater than 0.710 as P(Z > 0.710) = 1 - P(Z < 0.710) = 1 - 0.7611 = 0.2389.

Therefore, the probability that there are more passengers showing up for the flight than seats available, with the continuity correction, is approximately 0.2389.

Explanation: